Taylor series with a base point different from $0$

Question

What's the need for $f(x) = \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x-a)^{k}$ if we already have the formula at $0$?
Isn't the $(x-a)$ just making the $a$ as the new origin?
When is this formula more useful than that at $0$?

Answer 1

Some functions, like $\frac1x$, or $\ln x$, do not have a Taylor series centered around $0$. Around $1$, on the other hand, we have $$ \frac1x = 1-(x-1) + (x-1)^2 - (x-1)^3 + \cdots\\ \ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots $$

Some times, you're interested in approximating a function value at some point $x_0$, which is far from $0$, but close to some other point $a$ where you know the coefficients. For instance, finding $\sin(6.27)$ using the Taylor series of $\sin$ centered at $2\pi$ is going to go much faster than using the one centered at $0$.

Answer 2

If you are going to use Taylor polynomials to compute, say, an approximation of $\sqrt{4+\frac15}$, what you'll need is the Taylor series of $\sqrt x$ centered at $4$, not centered at $0$ (which, by the way, doesn't exist, since $\sqrt x$ isn't even differentiable there).